Question

A thin glass rod is bent into semicircle of radius r. A charge +Q is uniformly distributed along the upper half and a charge -Q is uniformly distributed along the lower half, as shown in figure. The electric field E at P, the centre of the semicircle, is

Answer 1

Correct option: (a)

Explanation: 

Take PQ as the x - axis and PA as the y - axis. Consider two elements EF and E’F’ of width dθ at angular distance θ at angular distance θ above and below PO, respectively The magnitude of the filed at P due to either element is  

Resolving the fields, we find that the components along PO sum up to zero and hence, the resultant field is along PB. 

Therefore, field at P due to pair of elements = 2d E sin θ